Statement-I: Polar solvent slows down `S_(N^(2))` reaction.
Because Statement-II: `CH_(3)-Br` is less reactive than `CH_(3)CI`.

Answer

To analyze the given statements, we will break down the information provided in each statement and evaluate their validity step by step.

### Step 1: Evaluate Statement-I
**Statement-I:** Polar solvent slows down `S_(N^(2))` reaction.

**Explanation:**
- In an `S_(N^(2))` reaction, a nucleophile attacks an electrophile (the carbon atom bonded to the leaving group).
- Polar solvents can solvate the nucleophile, which means they surround the nucleophile with solvent molecules.
- This solvation stabilizes the nucleophile and makes it less reactive, as it cannot effectively attack the electrophile.
- Therefore, the presence of a polar solvent indeed slows down the `S_(N^(2))` reaction.

**Conclusion for Statement-I:** True

### Step 2: Evaluate Statement-II
**Statement-II:** `CH_(3)-Br` is less reactive than `CH_(3)-Cl`.

**Explanation:**
- Reactivity in `S_(N^(2))` reactions is influenced by the bond dissociation energy (BDE) of the carbon-halogen bond.
- The bond between carbon and bromine (C-Br) in `CH3Br` is weaker than the bond between carbon and chlorine (C-Cl) in `CH3Cl`.
- This means that `CH3Br` will break apart more easily than `CH3Cl`, making `CH3Br` more reactive in nucleophilic substitution reactions.
- Therefore, the statement that `CH3Br` is less reactive than `CH3Cl` is incorrect.

**Conclusion for Statement-II:** False

### Final Conclusion
- Statement-I is true, and Statement-II is false.

### Summary of Results:
- **Statement-I:** True
- **Statement-II:** False